Answer: The answer is 50980
Step by Step Explanation:
First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.
